An important and exam oriented question from arithmetic progression chapter in which we are to find the AP, if the sum of the first 7 terms of an AP is 182 and its 4th and 17th terms are in the ratio 1:5.
Book RS Aggarwal, Class 10, chapter 5C, question no 27.
S7=7/2[2a+(7−1)d]=182
182=7/2[2a+6d]……..(1)
an=a+(n−1)d
a4=a+(4−1)d=a+3d
a17=a+(17−1)d=a+16d
From given, we have,
a4/a17=(a+3d)/(a+16d)
1/5=(a+3d)/(a+16d)
5a+15d=a+16d
solving the above equation, we get,
4a=d
using this value in (1), we get,
182=7/2[2a+6(4a)]
solving the above equation, we get,
a=2
4(2)=8=d
The series as follows,
a,a+d,a+2d,.........
⇒2,10,18,.......